Question

If $sec\theta +tan\theta =k,cos\theta =$

• A. $\frac{k^2+1}{2k}$
• B. $\frac{2k}{k^2+1}$
• C. $\frac{k}{k^2+1}$
• D. $\frac{k}{k^2-1}$

Answer 1

The correct option is B

$\frac{2k}{k^2+1}$

$\frac{2}{k^2+1}$

We have:

$sec\theta +tan\theta =k....\left(1\right)$

$\Rightarrow \frac{1}{sec\theta +tan\theta }=\frac{1}{k}$

$\Rightarrow \frac{se{c}^2\theta -ta{n}^2\theta }{sec\theta +tan\theta }=\frac{1}{k}$

$\Rightarrow \frac{(sec\theta +tan\theta )(sec\theta -tan\theta )}{(sec\theta +tan\theta )}=\frac{1}{k}$

$\therefore sec\theta -tan\theta =\frac{1}{k}......\left(2\right)$

Adding (1)and (2):

$2sec\theta =k+\frac{1}{k}$

$\Rightarrow 2sec\theta =\frac{k^2+1}{k}$

$\Rightarrow sec\theta =\frac{k^2+1}{2k}$

$\Rightarrow \frac{1}{cos\theta }=\frac{k^2+1}{2k}$

$\Rightarrow cos\theta =\frac{2k}{k^2+1}$